How to calculate partial derivative of an unknown function

Question

How can I force Mathematica to calculate symbolically the partial derivative of a function u[x,y] with respect to a variable z = f(x, y), where f(x, y) is known?

u is a function, defined on reals, taking real values. Same thing holds for f.

I want to try different changes of variables in PDEs. An example of what I'm trying to achieve and why it's failing:

z = 2*x + y 

2 x + y

D[u[x, y], z]

General::ivar: 2 x+y is not a valid variable. >>

D[u[x, y], 2*x + y]

EDIT: It appears I have failed to convey the essence of the question.

I am not asking how to solve a PDE with Mathematica. I am asking how, given a transformation of the original variables, to calculate the partial derivatives with respect to the new variables. The PDE part is merely some context as to why I want to do this.

Answer 1

In addition to belisarius' concrete answer, here is a more symbolic formulation:

If u and v are the new variables and the transformation functions are known to be

{x[u, v], y[u, v]}

then the set of partial derivatives of f with respect to u and v (i.e., the gradient) is

D[f[x, y], {{x, y}}].D[{x[u, v], y[u, v]}, {{u, v}}]

$\left\{ \\ f^{(1,0)}(x,y) x^{(1,0)}(u,v)+f^{(0,1)}(x,y)y^{(1,0)}(u,v), \\ f^{(1,0)}(x,y) x^{(0,1)}(u,v)+f^{(0,1)}(x,y)y^{(0,1)}(u,v)\\ \right\}$

Answer 2

Not sure if this is what you want. Anyway:

Suppose we define new coordinates:

{w == x + y, z == x - y}

Now:

ClearAll[f];
sol = Solve[{w == x + y, z == x - y}, {x, y}];
FullSimplify@D[f[x, y] /. sol[[1]], z]

Let's try it:

f[x_, y_] := (x + y)^2
FullSimplify@D[f[x, y] /. sol[[1]], z]
(* 0 *)
FullSimplify@D[f[x, y] /. sol[[1]], w]
(* 2 w *)

Answer 3

You need to use Dt.

Dt[u[x, y], f[x, y]]

(* Dt[y, f[x, y]]*Derivative[0, 1][u][x, y] + Dt[x, f[x, y]]*Derivative[1, 0][u][x, y] *)